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Solved Examples Chemistry | 11.43 Solved Examples JEE Main/Boards Example 1: Complete the following reactions: NBS Alc.KOH (B) (C ) (a) Me (A) Me Br Alc.KOH (b) (Y) + (Z) (X) (Major) (Minor) Me H (c ) (B) (A) Cl 1. Li -B 2 (B) (C ) (D) (d) Me at 500oC 2. CuI (A) (e) HO KHSO 4 (Y) OH or (X) Al23O or PO25 Allylic Br Br 3 1 3 NBS Alc. KOH Sol: (a) 4 4 Me Allylic Me 2 -Hbr 2 But-1-ene Substitution 3-Bromo-but-1-ene Buta-1,3-diene (A) (b) (Z) is formed by Saytzeff’s elimination, but the product (Y) is formed in major amount because product (Y) is a more stable conjugate diene than (Z), isolated diene 6 Me 4 2 -HBr Br 1 Me6 H at C-3 5 3 4 2 removed Me (b) 5 3 1 5-methly hexa-1,3-diene Me H H Alk.KOH (Y) Major 4-Bromo-5-methylhex-1-ene Me6 (X) 4 2 -HBr 5 1 H at C-3 3 is removed Me Saytzeff 5-methly hexa-1,4-diene elimination (Z) Minor 11.44 | Alkenes and Alkynes (c) 4 1 H H 3 5 -H 2 6 ( c ) H H 2 6 3 5 1 4 Less stable More stable isolated diene conjugated diene (A) (B) Cyclohexa-1,4-diene Cyclohexa-1,3-diene 2 o 2 Cl at 500 2 4 2 3 1. Li (d) 1 1 1 Me Allylic 2. Cu 3 (d) Prop-1-ene substitution Cl Corey-House Allylic synthesis Diallyl R (A) 3-Chloro- lithium cuperate prop-1-ene (B) Cl 4 4 6 1 3 5 R’ H (e) (e) HO 2 4 KHSO4 OH Or 1 3 H Al23O or PO25 (X) (Y) (-H2O) Butan-1,4-diol Butan-1,3-diene Example 2: Give the major products (not stereoisomers) of the following: 5 5 (a 1 4 (a)(a) 1 4 22 33 Cyclopenta-1,3-diene 1 1mo mol l 1 mo1 mol l 1 mo1l mol 11 mol of BrBr22 HBrHBr + + HBHBr r HS2 +HS + peroxide 2 peroxide peroperoxide xide (b) 55 (b) 1 (b) Me 1 4 Me 4 2 3 3 1-Ethylcyclopenta-1,3-diene2 1-Ethylcyclopenta-1,3-diene 1 mol 1 mol 1 mol 1 mol of Br2 1 mol of Br HBr1 mo + l HBr1 mol 1 mol 2 HS2 + peroHBrxide + HBr peroHSxide2 + peroxide peroxide Chemistry | 11.45 Sol: 1 (a) (a) 2 5 3 4 (A) All 1,4-addition products HBr or HS+ Br2 2 1,4-Add. HBr+peroxide Peroxide Br H H 3 5 2 4 1 4 1 4 1 2 2 5 Br 3 Br 3 SH 3,5-Dibromo- 3-Bromo- Cyclopent-1- cyclopent-1-ene cyclopent-1-ene ene-3-thiol (b) 1 (b) Me 4 2 3 All 1,4-addition products HBr + peroxide HBr + peroxide HS + peroxide 1 mol of Br2 2 Anti-Markonvikov’s Anti-Markonvikov’s Anti-Markonvikov’s add. add. add. 4 4 4 4 1 5 Br 3 5 Br 1 5 H 5 3 SH Me Br Me H Me Br Me H 2 3 2 1 2 3 1 2 3,5-Dibromo- 5-Bromo-3-ethyl- 3-Bromo- 5-Ethyl cyclo- 3-ethyl-cyclopent-1-ene cyclopent-1-ene 3-ethyl-cyclopent-1-ene pent-1-ene-3-thiol Example 3: Write all the possible structures and give the structure of the products that are thermodynamically favoured.2 4 6 1 mol of 2 4 Products 1 3 Br6 Sol: 5 2 1 mol of 1 Products Hexa-1,3-5-triene3 5 Br2 Hexa-1,3-5-triene 2 4 6 1 3 2 5 4 6 Br Br 1 2 3 5 Br Br Br2 Br 2 Br 2 6 6 4 4 Br 1 3 5 1 3 5 2 1,4-AdditionBr 2 6 Br 6 4 4 Br 4 5 Br 1 3 6 5 Br 1 4 3 2 Br 1,4-Addition6 1 3 5 2 4 1 3 Br5 Br 6 Br 1,2-Addition 4 2 4 Br 6 Br 2 2 4 6 1 3 6 5 Br 2 1 3 5 1 3 Br 5 1 3 1,2-Addition5 (III) 2 4 1,6-Addition Br 2 4 6 6 Br 1 3 5 Br 5 Br 2 4 Br 2 1 3 (III) 6 2 1,6-Addition6 Addition of 4 1 3 5 1 3 5 Br23at C-C4 Br 2 4 double bond 2 6 Br2 6 Addition of 4 1 3 5 1 3 5 Br23at C-C4 double bond 2 4 6 1 mol of 1 Products 3 5 Br2 Hexa-1,3-5-triene 2 4 6 1 3 5 Br Br2 Br Br 2 Br 2 6 6 4 4 1 3 5 1 3 5 1,4-Addition 4 Br Br 6 Br 4 2 Br 6 1 3 5 2 1 3 5 1,2-Addition 2 4 Br 2 4 6 6 Br 1 3 5 Br 11.46 | Alkenes and Alkynes 1 3 5 (III) 1,6-Addition Br 2 4 2 6 Br2 6 Addition of 4 1 3 5 1 3 5 Br23at C-C4 double bond More stable Products (II and III, conjugate dienes) are thermodynamically2 favoured4 products than2 I and 4IV (isolated dienes) (a) and 1 3 5 1 3 5 (A) (B) Example 4: Distinguish between the given pairs. 3 2 4 2 4 5 4 3 2 1 2 C = C=C COOH and (a) and (b) 6 4 1 1 3 5 1 3 5 H H COOH (A) (B) (C ) (D) Sol: (a) Compounds (A) and (B) are distinguished3 chemically by quantitative catalytic hydrogenation. Since 5 4 3 2 1 2 compound (A) has two doubleand bond whereas compound (B) has only one double bond .Compound (A) will require 6 C = C=C COOH 4 (b)2 moles of H for 1 mole of (A) while (B) will1 requires 1 mole of H per mole of (B). H2 H COOH 2 ( ) (D) C COOH Me C =C = C COOH H (C ) (b) Allene (C) with different groups on each of the double bonds is optically active and can be resolved into its enantiomers. Compound (D) is conjugate diene and does not show optical isomerism. Example 5: Give the major and minor products. CH CH2 H 2 1 CH2 3 2 6 HBr 2 4 o 1 mol HBr at 45 C 3 o 5 at 45 C Br 1 5 4 6 (B) (major) (A) 3-Bromo-1-methyl- 3-Methylenecyclohex-1-ene cyclohex-1-ene Sol: At high temperature (45ºC), 1, 4-addition products is more favourable. CH 2 Br CH CH2 H CH 2 1 2 1 2 6 2 6 HBr HBr o at 45 C 3 5 3 5 Br H 4 4 (A) (B) (major) (C ) (major) 3-Bromo-1-methyl- 1-Bromo-1-methyl- cyclohex-1-ene cyclohex-1-ene Example 6: Dehydration of (A) with conc.CH2 H BrSO gives a compound that exists in two CH 2 4 Me H isomeric forms.2 Give the structures of both1 the isomers. 2 6 H OH HBr (A) 3 5 H (A) 4 (C ) (major) 1-Bromo-1-methyl- cyclohex-1-ene Chemistry | 11.47 Sol: 5 6 Me H Conc. Me 1 H Me H 4 + H OH HS24O H H 3 (H) H H H (Z) or cis (E) or trans 4-Methylclohex-1-ene Example 7: Me H+ Lindlar’s Catalyst 2 (B) HS24O (A) Me Sol: Addition of H atom by Lindlar’s catalyst is a syn-addition. It reduces (C ≡ C) bond to (C = C) bond so the product (B) is H H O/3 Reduction Example 8: CH10 16(A) CH58(B) Not-reslovable Not-resolvable aldehyde O/3 Reduction CH58O2(C ) Not-resolvable acid 3 equivalent of H+2 Pd at 120oC (D) Write all the possible structures of (A), (B) and (C). Sol: Since the reductive and oxidative ozonolysis products are different, the alkene of the type HH is possible RR If , the possible structures of (A) can be: R = ( — C H2 —) , the possible structures of (A) can be : R = ( CH2 ( 5 6 H H H 3 mol 4 7 3 8 of H2 H 2 9 1 (D) 10 Both optically inactive Decane 4 5 O/3 Reduction O3/Oxidation 3 6 2 CH = O 2 COOH Me 2 7 Me (B) (B) 1 8 Optically inactive Optically inactive (D1) 2,7-Dimethyloctane 11.48 | Alkenes and Alkynes 2 1 (D) is obtained when (C1 – C3) bond breaks H ( H 3 ( 2 1 (D1) is obtained when (C2 – C3) bond H breaks during hydrogenation of (A). H ( 3 ( Example 9: Give the number of stereoisomers of (A) in the following reactions. O3 /oxidation (a) (A) (C7 H 12 ) Acetone + Oxalic acid + Acetic acid (B) (C) (D) O3 /oxidation (b) (A) (C8 H 14 ) Butane-2-one + Oxalic acid + Acetic acid (B) (C) (D) Sol: (a) Me O HOOC - COOH HOOC - Me Me (B) (C ) (D) Convert acid to aldehyde Me OO==HC - CH OO= HC - Me Me Remove and join the double bond to obtain the structure of (A) Me 1 23456 CH CH = CH Me Me (A) (C H) 712 6 H H Me Here Compound (A) shows G.I around C — C H 5 4 5 4 5 4 double bond (since the two groups around C-4 C C and C C 3 6 3 and C-5 are different) but do not show G.I around CH Me CH H Me1 2 Me1 2 C2 — C3 double bond because the two groups around C-2 are the same.
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